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PROPOSITION XXXIV.

THEOREM.

The opposite sides and angles of a parallelogram are equal to one another, agonal bisects it.

and

and

Let ABDC be a O, and BC a diagonal of the ].
Then must AB=DC and AC=DB,

L BAC=1 CDB, and ▲ ABD= 2 ACD
▲ ABC= ▲ DCB.

AB is | to CD, and BC meets them,
.. ABC alternate 4 DCB;

and AC is to BD, and BC meets them,

.. 4 ACB alternate DBC.

For

D

I. 29.

I. 29.

Then in as ABC, DCB,

;

· ▲ ABC= 2 DCB, and ▲ ACB= 4 DBC, and BC is common, a side adjacent to the equals in each .. AB=DC, and AC=DB, and ▲ BAC=▲ CDB, and ▲ ABC= ▲ DCB.

I. B.

Also

ABC= 4 DCB, and ▲ DBC= 2 ACB,

.. ¿s ABC, DBC together= 2 s DCB, ACB together, that is, L ABDL ACD.

Q. E. D.

Ex. 1. Shew that the diagonals of a parallelogram bisect each other.

Ex. 2. Shew that the diagonals of a rectangle are equal.

PROPOSITION XXXV.

THEOREM.

Parallelograms on the same base and between the same parallels are equal.

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Let the s ABCD, EBCF be on the same base BC and between the same ||s AF, BC.

Then must ABCD=☐ EBCF.

CASE I. If AD, EF have no point common to both,
Then in the AS FDC, EAB,

extr. ▲ FDC=intr. ▲ EAB, and intr. ▲ DFC=extr. ▲ AEB, and DC=AB,

:: ΔFDC= ΔΕΑΒ.

Now

ABCD with ▲ FDC=figure ABCF ; and ☐ EBCF with ▲ EAB=figure ABCF;

:.

ABCD with ▲ FDC=□ EBCF with ▲ EAB;
ABCD=□ EBCF.

I. 29.

I. 29.

I. 34.

I. 26.

the same method of proof applies.

CASE II. If the sides AD, EF overlap one another,

W

B

CASE III. If the sides opposite to BC be terminated in the same point D,

W

B

the same method of proof is applicable,
but it is easier to reason thus:

Each of the s is double of ▲ BDC;
.. ABCD= DBCF.

Then

THEOREM.

PROPOSITION XXXVI. Parallelograms on equal bases, and between the same parallels, are equal to one another.

A

D E

I. 34.

B

F

G

Let the Os ABCD, EFGH be on equal bases BC, FG, and between the same Is AH, BG.

Then must

ABCD=☐ EFGH.

CH.

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Q. E. D.

Join BE,

::: BC=FG,

and EH-FG;

.. BC=EH ; and BC is to EH. .. EB is || to CH;

.. EBCH is a parallelogram. Now EBCH=□ABCD,

I. 35.

• they are on the same base BC and between the same ||s; EBCH=□EFGH,

I. 35.

and ..they are on the same base EH and between the same ||s; ..ABCDEFGH.

Hyp.

I. 34.

Hyp.

I. 33.

Q. E. D.

PROPOSITION XXXVII. THEOREM.

Triangles upon the same base, and between the same parallels, are equal to one another.

B

Let As ABC, DBC be on the same base BC and between the same s AD, BC.

Then must ▲ ABC= ▲ DBC.

From B draw BE || to CA to meet DA produced in E.
From C draw CF || to BD to meet AD produced in F.
Then EBCA and FCBD are parallelograms,

and

EBCA=□ FCBD,

they are on the same base and between the same ||s. Now ▲ ABC is half of

EBCA,

and DBC is half of

FCBD;

.... AABC= ADBC.

I. 35.

I. 34.

I. 34.

Ax. 7.

Q. E. D.

Ex. 1. If P be a point in a side AB of a parallelogram ABCD, and PC, PD be joined, the triangles PAD, PBC are together equal to the triangle PDC.

Ex. 2. Two straight lines AB, CD intersect in E, and the triangle AEC is equal to the triangle BED. Shew that BC is parallel to AD.

Ex. 3. If A, B be points in one, and C, D points in another of two parallel straight lines, and the lines AD, BC intersect in E, then the triangles AEC, BED are equal.

PROPOSITION XXXVIII. THEOREM.

Triangles upon equal bases, and between the same parallels, are equal to one another.

B

AN

Let As ABC, DEF be on equal bases, BC, EF, and between the same ||s BF, AD.

Then must ▲ ABC= A DEF.

From B draw BG || to CA to meet DA produced in G. From F draw FH || to ED to meet AD produced in H. Then CG and EH are parallelograms, and they are equal, they are on equal bases BC, EF, and between the same Is BF, GH. I. 36.

Now AABC is half of
and A DEF is half of

. A ABC= ADEF.

CG,

EH;

Ax. 7.

Q. E. D.

Ex. 1. Shew that a straight line, drawn from the vertex of a triangle to bisect the base, divides the triangle into two equal parts.

Ex. 2. If the triangles in the Proposition are not towards the same parts, shew that the straight line, joining the vertices of the triangles, is bisected by the line containing the bases.

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Ex. 3. In the equal sides AB, AC of an isosceles triangle ABC points D, E are taken such that BD=AE. Shew that the triangles CBD, ABE are equal.

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